# Thread: i forget a very simple integration problem

1. ## i forget a very simple integration problem

i forget a very simple integration problem: is there any way to do this:

2. $\displaystyle \int {\sec \theta \,d\theta } = \int {\frac{{\cos \theta }} {{(1 + \sin \theta )(1 - \sin \theta )}}\,d\theta } = \frac{1} {2}\left( {\int {\frac{{\cos \theta }} {{1 + \sin \theta }}\,d\theta } + \int {\frac{{\cos \theta }} {{1 - \sin \theta }}\,d\theta } } \right).$

Or just multiply top & bottom by $\displaystyle \sec\theta+\tan\theta.$

3. Hello, afeasfaerw!

$\displaystyle \int \sec\theta\,d\theta$

You should have been taught the formula for this by now . . .

It is usually derived like this . . .

Multiply by .$\displaystyle \frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}$

. . $\displaystyle \int\frac{\sec\theta}{1}\cdot\frac{\sec\theta + \tan\theta}{\sec\theta + \tan\theta}\,d\theta \;=\;\int\frac{\sec^2\!\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\,d\theta \;=\;\int\frac{\sec\theta\tan\theta + \sec^2\!\theta}{\sec\theta + \tan\theta}\,d\theta$

Let: $\displaystyle u \:=\:\sec\theta + \tan\theta\quad\Rightarrow\quad du \:=\:(\sec\theta\tan\theta + \sec^2\theta)d\theta$

Substitute: .$\displaystyle \int\frac{du}{u}\;=\;\ln|u| + C$

Back-substitute: .$\displaystyle \ln|\sec\theta + \tan\theta| + C$